01 SDOF - SPC408 - Fall2016
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Transcript of 01 SDOF - SPC408 - Fall2016
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Single Degree of Freedom SystemsMaged Mostafa
Dynamics of Aerospace Structures
SPC408
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Single degree of freedom systems
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Objectives• Recognize a SDOF system• Be able to solve the free vibration equation
of a SDOF system with and without damping
• Understand the effect of damping on the system vibration
• Apply numerical tools to obtain the time response of a SDOF system
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Single degree of freedom systems
• When one variable can describe the motion of a structure or a system of bodies, then we may call the system a 1-D system or a single degree of freedom (SDOF) system. e.g. x(t), q(t) Z(t), y(x).
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Stiffness
• From strength of materials recall:
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Newton’s Law
• Newton’s Law:
00 )0(,)0(0)()(
)()(
vxxxtkxtxm
tkxtxm
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Solving the ODE
• The ODE is• The proposed solution:• Into the ODE you get the characteristic
equation:
• Giving:
0)()( tkxtxm taetx )(
02 tt aemkae
mk
2mkj
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Solving the ODE (cont’d)
• The proposed solution becomes:
• For simplicity, let’s define:
• Giving:
tmkjt
mkj
eaeatx
21)(
mk
tjtj eaeatx 21)(
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Let’s manipulate the solution!
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Recall
ajSinaCose ja
bSinaCosbCosaSinbaSin
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Manipulating the solution
• The solution we have:• Rewriting:
tjtj eaeatx 21)(
tjSintCosatjSintCosatx
2
1)(
tSinaajtCosaatx 2121)(
tSinAtCosAtx 21)(
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Further manipulation tSinAtCosAtx 21)(
22
21 AAA
AASin
AACos 12 &
tSinCostCosSinAtx )(
tASintx )(
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Different forms of the solution
tjtj eaeatx
tCosAtSinAtx
tASintx
21
21
)(
)(
)()(
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NOTE!
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Natural Frequency of Oscillation• In the previously obtained solution:
• The frequency of oscillation is • It depends only on the characteristics of the
oscillating system. That is why it is called the natural frequency of oscillation
tASintx )(
mk
n
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Frequency
period theis s 2
Hz2s 2
cycles rad/cycle 2
rad/s frequency natural thecalled is rad/sin is
n
nnnn
n
T
f
We often speak of frequency in Hertz or RPM, but we need rad/s in the arguments of the trigonometric functions.
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Recall: Initial Conditions
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Amplitude & Phase from the ICs
Phase
0
01
Amplitude
2
202
0
0
0
tan ,
yields Solvingcos)0cos(
sin)0sin(
vxvxA
AAvAAx
n
n
nnn
n
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Some useful quantities
peak value A
T
Tdttx
Tx
0
valueaverage = )(1lim
valuesquaremean root = 2xxrms
valuesquare-mean = )(1lim0
22
T
Tdttx
Tx
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Peak Values
Ax
AxAx
2max
max
max
:onaccelerati
:velocity :ntdisplaceme
Maximum or peak (amplitude) values:
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Samples of Vibrating Systems
• Deflection of continuum (beams, plates, bars, etc) such as airplane wings, truck chassis, disc drives, circuit boards…
• Shaft rotation• Rolling ships
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Wing Vibration
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Ship Vibration
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Effective Stiffness of Structures
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Bars• Longitudinal motion• A is the cross sectional
area (m2)• E is the elastic modulus
(Pa=N/m2)• l is the length (m)• k is the stiffness (N/m) x(t)
m
EAk
l
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Rods
• Jp is the polar moment of inertia of the rod
• J is the mass moment of inertia of the disk
• G is the shear modulus, l is the length
Jp
J qt)
0
pGJk
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Helical Spring
2R
x(t)
d = diameter of wire2R= diameter of turns n = number of turns x(t)= end deflectionG= shear modulus of spring material
3
4
64nRGdk
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Beams
f
m
x
• Strength of materials and experiments yield:
3
3
3
3
mEI
EIk
n
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Equivalent Stiffness
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Example 1.5.2 Effect of fuel on frequency of an airplane wing
• Model wing as transverse beam
• Model fuel as tip mass• Ignore the mass of the
wing and see how the frequency of the system changes as the fuel is used up
x(t) l
E, I m
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Mass of pod 10 kg empty 1000 kg fullI = 5.2x10-5 m4, E =6.9x109 N/m, l = 2 m
• Hence the natural frequency changes by an order of magnitude while it empties out fuel.
Hz 18.5=rad/s 115 210
)102.5)(109.6(33
Hz 1.8=rad/s 6.11 21000
)102.5)(109.6(33
3
59
3empty
3
59
3full
mEI
mEI
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Example 1.5.5 Design of a spring mass system using available springs: series vs parallel
• Let m = 10 kg• Compare a series and
parallel combination• a) k1 =1000 N/m, k2 = 3000
N/m, k3 = k4 =0• b) k3 =1000 N/m, k4 = 3000
N/m, k1 = k2 =0
k1 k2
k3 k4
m
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rad/s 66.810750
N/m 75013
3000)1()1(
1,0
:connection series b) Case
rad/s 2010
4000
N/m 400030001000,0:connection parallel a) Case
4321
2143
mk
kkkkk
mk
kkkkk
egseries
eq
egparallel
eq
Same physical components, very different frequencyAllows some design flexibility in using off the shelf components
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Harmonic Excitation of Undamped Systems
• Consider the usual spring mass damper system with applied force F(t)=F0cost
• is the driving frequency• F0 is the magnitude of the applied force• We take c = 0 to start with
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Equation of motion
mkmFf
tftxtx
tFtkxtxm
n
n
,/ where
)cos()()(
)cos()()(
00
02
0
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Linear non-homogenous ODE:
• Solution is sum of homogenous and particular solution
• The particular solution assumes a form of forcing function (physically the input wins)
)cos()( tXtxp
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Substitute into the equation of motion:
220
022
:yields solving
coscoscos
2
n
x
n
x
fX
tftXtXpnp
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Thus the particular solution has the form:
)cos()( 220 tftx
np
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General Solution
particular
nnn tftAtAtx
coscossin )( 22
0
shomogeneou
21
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01
0220
2
)0(
)0(
vAx
xfAx
n
n
Apply the initial conditions to evaluate the constants
Solving for the constants and substituting into x yields
tftfxtvtxn
nn
nn
coscossin)( 220
220
00
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Comparison of free and forced response• Sum of two harmonic terms of different
frequency• Free response has amplitude and phase
affected by forcing function• Our solution is not defined for n =
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Example• Compute and plot the response for m=10 kg, k=1000 N/m,
x0=0,v0=0.2 m/s, F=23 N, =2n
rad/s 202 rad/s, 10kg 10N/m 1000
nn mk
Solution
m 02.0rad/s 10m/s 2.0 N/kg, 3.2
kg 10N 23 0
0 n
vmFf
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Example (cont’d)
m 109667.7s/rad )20(10
N/kg 3.2 3222222
0
n
f
)20cos10(cos10667.710sin02.0)( 3 ttttx
Substituting into the general solution:
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Response to Harmonic Excitation
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Beat !
• What happens when the driving frequency is near the natural frequency?
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Example• Given zero initial conditions a harmonic input of 10 Hz
with 20 N magnitude and k= 2000 N/m, and measured response amplitude of 0.1m, compute the mass of the system.
ttftx nn
coscos)( 220
Solution:
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Example (cont’d)
• Using Trigonometric Identities:
• Using given data
termFast
n
termSlow
n
n
ttftx
2sin
2sin2)( 22
0
kg 45.0
1.0)20(2000
)/20(21.02222
0
mm
mf
n
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Plotting the result
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Twice Frequency and Beat
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Resonance
• When excitation frequency is equal to the natural frequency, the previous solution fails.
• The particular solution of the ODE become:
)sin()( ttXtxp
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Substituting!
20fX
Use the particular solution into the equation:
bound without grows
021 )sin(
2cossin)( ttftAtAtx
The total solution becomes:
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Resonance
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Summary• Write down the equation of motion using Newton’s law• Solve the equation of motion for a SDOF• Use initial conditions to determine the amplitude and
phase of vibration for a SDOF• Evaluate the effective stiffness of structural members• Analyzing the response of a SDOF to harmonic
excitations• Different types of response depend on the excitation
frequency• What happens during the beat and resonance
phenomena
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1. The amplitude of vibration of an undamped system is measured to be 1 mm. the phase shift is measured to be 2 rad and the frequency 5 rad/sec. Calculate the initial conditions.
2. Using the equation:
evaluate the constant A1 and A2 in terms of the initial conditions
HW #1
tSinAtCosAtx 21)(
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HW #1 (cont’d)
3. A vehicle is modeled as 1000 kg mass supported by a stiffness k=400 kN/m. When it oscillates, the maximum deflection is 10 cm. when loaded with the passengers, the mass becomes 1300 kg. calculate the change in the frequency, velocity amplitude, and acceleration if the maximum deflection remain 10 cm.